3.55.54 \(\int \frac {-6144 x+3840 x^3-80 x^4-600 x^5+15 x^6+(768 x^3-240 x^5+3 x^6) \log (3)-24 x^5 \log ^2(3)+e^x (-2048 x-1024 x^2+1280 x^3+640 x^4-200 x^5-100 x^6+(256 x^3+128 x^4-80 x^5-40 x^6) \log (3)+(-8 x^5-4 x^6) \log ^2(3))}{1024-640 x^2+100 x^4+(-128 x^2+40 x^4) \log (3)+4 x^4 \log ^2(3)} \, dx\)

Optimal. Leaf size=27 \[ x^2 \left (-3-e^x+\frac {x}{4 \left (5-\frac {16}{x^2}+\log (3)\right )}\right ) \]

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Rubi [B]  time = 1.28, antiderivative size = 71, normalized size of antiderivative = 2.63, number of steps used = 27, number of rules used = 14, integrand size = 163, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6, 6688, 12, 6742, 2196, 2176, 2194, 261, 288, 321, 206, 266, 43, 302} \begin {gather*} \frac {5 x^3}{8 (5+\log (3))}-e^x x^2-3 x^2+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6144*x + 3840*x^3 - 80*x^4 - 600*x^5 + 15*x^6 + (768*x^3 - 240*x^5 + 3*x^6)*Log[3] - 24*x^5*Log[3]^2 + E
^x*(-2048*x - 1024*x^2 + 1280*x^3 + 640*x^4 - 200*x^5 - 100*x^6 + (256*x^3 + 128*x^4 - 80*x^5 - 40*x^6)*Log[3]
+ (-8*x^5 - 4*x^6)*Log[3]^2))/(1024 - 640*x^2 + 100*x^4 + (-128*x^2 + 40*x^4)*Log[3] + 4*x^4*Log[3]^2),x]

[Out]

-3*x^2 - E^x*x^2 + (5*x^3)/(8*(5 + Log[3])) + (3*x^5)/(8*(16 - x^2*(5 + Log[3]))) - (10*x^3)/((5 + Log[3])*(16
- x^2*(5 + Log[3])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6144 x+3840 x^3-80 x^4-600 x^5+15 x^6+\left (768 x^3-240 x^5+3 x^6\right ) \log (3)-24 x^5 \log ^2(3)+e^x \left (-2048 x-1024 x^2+1280 x^3+640 x^4-200 x^5-100 x^6+\left (256 x^3+128 x^4-80 x^5-40 x^6\right ) \log (3)+\left (-8 x^5-4 x^6\right ) \log ^2(3)\right )}{1024-640 x^2+\left (-128 x^2+40 x^4\right ) \log (3)+x^4 \left (100+4 \log ^2(3)\right )} \, dx\\ &=\int \frac {-6144 x+3840 x^3-80 x^4+15 x^6+\left (768 x^3-240 x^5+3 x^6\right ) \log (3)+x^5 \left (-600-24 \log ^2(3)\right )+e^x \left (-2048 x-1024 x^2+1280 x^3+640 x^4-200 x^5-100 x^6+\left (256 x^3+128 x^4-80 x^5-40 x^6\right ) \log (3)+\left (-8 x^5-4 x^6\right ) \log ^2(3)\right )}{1024-640 x^2+\left (-128 x^2+40 x^4\right ) \log (3)+x^4 \left (100+4 \log ^2(3)\right )} \, dx\\ &=\int \frac {x \left (-6144-80 x^3+768 x^2 (5+\log (3))+3 x^5 (5+\log (3))-24 x^4 (5+\log (3))^2-4 e^x (2+x) \left (-16+x^2 (5+\log (3))\right )^2\right )}{4 \left (16-x^2 (5+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {x \left (-6144-80 x^3+768 x^2 (5+\log (3))+3 x^5 (5+\log (3))-24 x^4 (5+\log (3))^2-4 e^x (2+x) \left (-16+x^2 (5+\log (3))\right )^2\right )}{\left (16-x^2 (5+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-4 e^x x (2+x)-\frac {6144 x}{\left (16-x^2 (5+\log (3))\right )^2}-\frac {80 x^4}{\left (16-x^2 (5+\log (3))\right )^2}+\frac {768 x^3 (5+\log (3))}{\left (16-x^2 (5+\log (3))\right )^2}+\frac {3 x^6 (5+\log (3))}{\left (16-x^2 (5+\log (3))\right )^2}-\frac {24 x^5 (5+\log (3))^2}{\left (16-x^2 (5+\log (3))\right )^2}\right ) \, dx\\ &=-\left (20 \int \frac {x^4}{\left (16-x^2 (5+\log (3))\right )^2} \, dx\right )-1536 \int \frac {x}{\left (16-x^2 (5+\log (3))\right )^2} \, dx+\frac {1}{4} (3 (5+\log (3))) \int \frac {x^6}{\left (16-x^2 (5+\log (3))\right )^2} \, dx+(192 (5+\log (3))) \int \frac {x^3}{\left (16-x^2 (5+\log (3))\right )^2} \, dx-\left (6 (5+\log (3))^2\right ) \int \frac {x^5}{\left (16-x^2 (5+\log (3))\right )^2} \, dx-\int e^x x (2+x) \, dx\\ &=\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {768}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {15}{8} \int \frac {x^4}{16+x^2 (-5-\log (3))} \, dx+\frac {30 \int \frac {x^2}{16+x^2 (-5-\log (3))} \, dx}{5+\log (3)}+(96 (5+\log (3))) \operatorname {Subst}\left (\int \frac {x}{(16+x (-5-\log (3)))^2} \, dx,x,x^2\right )-\left (3 (5+\log (3))^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{(16+x (-5-\log (3)))^2} \, dx,x,x^2\right )-\int \left (2 e^x x+e^x x^2\right ) \, dx\\ &=-\frac {30 x}{(5+\log (3))^2}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {768}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-\frac {15}{8} \int \left (-\frac {16}{(5+\log (3))^2}-\frac {x^2}{5+\log (3)}+\frac {256}{(5+\log (3))^2 \left (16-x^2 (5+\log (3))\right )}\right ) \, dx-2 \int e^x x \, dx+\frac {480 \int \frac {1}{16+x^2 (-5-\log (3))} \, dx}{(5+\log (3))^2}+(96 (5+\log (3))) \operatorname {Subst}\left (\int \left (\frac {16}{(5+\log (3)) (16-x (5+\log (3)))^2}+\frac {1}{(-5-\log (3)) (16-x (5+\log (3)))}\right ) \, dx,x,x^2\right )-\left (3 (5+\log (3))^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{(5+\log (3))^2}+\frac {256}{(5+\log (3))^2 (16-x (5+\log (3)))^2}+\frac {32}{(5+\log (3))^2 (-16+x (5+\log (3)))}\right ) \, dx,x,x^2\right )-\int e^x x^2 \, dx\\ &=-2 e^x x-3 x^2-e^x x^2+\frac {120 \tanh ^{-1}\left (\frac {1}{4} x \sqrt {5+\log (3)}\right )}{(5+\log (3))^{5/2}}+\frac {5 x^3}{8 (5+\log (3))}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}+2 \int e^x \, dx+2 \int e^x x \, dx-\frac {480 \int \frac {1}{16-x^2 (5+\log (3))} \, dx}{(5+\log (3))^2}\\ &=2 e^x-3 x^2-e^x x^2+\frac {5 x^3}{8 (5+\log (3))}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}-2 \int e^x \, dx\\ &=-3 x^2-e^x x^2+\frac {5 x^3}{8 (5+\log (3))}+\frac {3 x^5}{8 \left (16-x^2 (5+\log (3))\right )}-\frac {10 x^3}{(5+\log (3)) \left (16-x^2 (5+\log (3))\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 48, normalized size = 1.78 \begin {gather*} \frac {x^2 \left (192+x^3-12 x^2 (5+\log (3))-4 e^x \left (-16+x^2 (5+\log (3))\right )\right )}{4 \left (-16+x^2 (5+\log (3))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6144*x + 3840*x^3 - 80*x^4 - 600*x^5 + 15*x^6 + (768*x^3 - 240*x^5 + 3*x^6)*Log[3] - 24*x^5*Log[3]
^2 + E^x*(-2048*x - 1024*x^2 + 1280*x^3 + 640*x^4 - 200*x^5 - 100*x^6 + (256*x^3 + 128*x^4 - 80*x^5 - 40*x^6)*
Log[3] + (-8*x^5 - 4*x^6)*Log[3]^2))/(1024 - 640*x^2 + 100*x^4 + (-128*x^2 + 40*x^4)*Log[3] + 4*x^4*Log[3]^2),
x]

[Out]

(x^2*(192 + x^3 - 12*x^2*(5 + Log[3]) - 4*E^x*(-16 + x^2*(5 + Log[3]))))/(4*(-16 + x^2*(5 + Log[3])))

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fricas [B]  time = 2.02, size = 59, normalized size = 2.19 \begin {gather*} \frac {x^{5} - 12 \, x^{4} \log \relax (3) - 60 \, x^{4} + 192 \, x^{2} - 4 \, {\left (x^{4} \log \relax (3) + 5 \, x^{4} - 16 \, x^{2}\right )} e^{x}}{4 \, {\left (x^{2} \log \relax (3) + 5 \, x^{2} - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^6-8*x^5)*log(3)^2+(-40*x^6-80*x^5+128*x^4+256*x^3)*log(3)-100*x^6-200*x^5+640*x^4+1280*x^3-1
024*x^2-2048*x)*exp(x)-24*x^5*log(3)^2+(3*x^6-240*x^5+768*x^3)*log(3)+15*x^6-600*x^5-80*x^4+3840*x^3-6144*x)/(
4*x^4*log(3)^2+(40*x^4-128*x^2)*log(3)+100*x^4-640*x^2+1024),x, algorithm="fricas")

[Out]

1/4*(x^5 - 12*x^4*log(3) - 60*x^4 + 192*x^2 - 4*(x^4*log(3) + 5*x^4 - 16*x^2)*e^x)/(x^2*log(3) + 5*x^2 - 16)

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giac [B]  time = 0.23, size = 63, normalized size = 2.33 \begin {gather*} -\frac {4 \, x^{4} e^{x} \log \relax (3) - x^{5} + 20 \, x^{4} e^{x} + 12 \, x^{4} \log \relax (3) + 60 \, x^{4} - 64 \, x^{2} e^{x} - 192 \, x^{2}}{4 \, {\left (x^{2} \log \relax (3) + 5 \, x^{2} - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^6-8*x^5)*log(3)^2+(-40*x^6-80*x^5+128*x^4+256*x^3)*log(3)-100*x^6-200*x^5+640*x^4+1280*x^3-1
024*x^2-2048*x)*exp(x)-24*x^5*log(3)^2+(3*x^6-240*x^5+768*x^3)*log(3)+15*x^6-600*x^5-80*x^4+3840*x^3-6144*x)/(
4*x^4*log(3)^2+(40*x^4-128*x^2)*log(3)+100*x^4-640*x^2+1024),x, algorithm="giac")

[Out]

-1/4*(4*x^4*e^x*log(3) - x^5 + 20*x^4*e^x + 12*x^4*log(3) + 60*x^4 - 64*x^2*e^x - 192*x^2)/(x^2*log(3) + 5*x^2
- 16)

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maple [B]  time = 0.32, size = 57, normalized size = 2.11

method	result	size
norman	\(\frac {\left (-3 \ln \relax (3)-15\right ) x^{4}+48 x^{2}+\left (-5-\ln \relax (3)\right ) x^{4} {\mathrm e}^{x}+\frac {x^{5}}{4}+16 \,{\mathrm e}^{x} x^{2}}{x^{2} \ln \relax (3)+5 x^{2}-16}\)	\(57\)
risch	\(-\frac {12 \ln \relax (3)^{2} x^{2}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {\ln \relax (3) x^{3}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}-\frac {120 \ln \relax (3) x^{2}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {5 x^{3}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}-\frac {300 x^{2}}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {16 x}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right )}+\frac {256 x}{\left (4 \ln \relax (3)+20\right ) \left (5+\ln \relax (3)\right ) \left (x^{2} \ln \relax (3)+5 x^{2}-16\right )}-{\mathrm e}^{x} x^{2}\)	\(160\)
default	\(\text {Expression too large to display}\)	\(3045\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^6-8*x^5)*ln(3)^2+(-40*x^6-80*x^5+128*x^4+256*x^3)*ln(3)-100*x^6-200*x^5+640*x^4+1280*x^3-1024*x^2-
2048*x)*exp(x)-24*x^5*ln(3)^2+(3*x^6-240*x^5+768*x^3)*ln(3)+15*x^6-600*x^5-80*x^4+3840*x^3-6144*x)/(4*x^4*ln(3
)^2+(40*x^4-128*x^2)*ln(3)+100*x^4-640*x^2+1024),x,method=_RETURNVERBOSE)

[Out]

((-3*ln(3)-15)*x^4+48*x^2+(-5-ln(3))*x^4*exp(x)+1/4*x^5+16*exp(x)*x^2)/(x^2*ln(3)+5*x^2-16)

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maxima [B]  time = 0.52, size = 848, normalized size = 31.41 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^6-8*x^5)*log(3)^2+(-40*x^6-80*x^5+128*x^4+256*x^3)*log(3)-100*x^6-200*x^5+640*x^4+1280*x^3-1
024*x^2-2048*x)*exp(x)-24*x^5*log(3)^2+(3*x^6-240*x^5+768*x^3)*log(3)+15*x^6-600*x^5-80*x^4+3840*x^3-6144*x)/(
4*x^4*log(3)^2+(40*x^4-128*x^2)*log(3)+100*x^4-640*x^2+1024),x, algorithm="maxima")

[Out]

-x^2*e^x - 3*(x^2/(log(3)^2 + 10*log(3) + 25) + 32*log(x^2*(log(3) + 5) - 16)/(log(3)^3 + 15*log(3)^2 + 75*log
(3) + 125) - 256/((log(3)^4 + 20*log(3)^3 + 150*log(3)^2 + 500*log(3) + 625)*x^2 - 16*log(3)^3 - 240*log(3)^2
- 1200*log(3) - 2000))*log(3)^2 - 30*(x^2/(log(3)^2 + 10*log(3) + 25) + 32*log(x^2*(log(3) + 5) - 16)/(log(3)^
3 + 15*log(3)^2 + 75*log(3) + 125) - 256/((log(3)^4 + 20*log(3)^3 + 150*log(3)^2 + 500*log(3) + 625)*x^2 - 16*
log(3)^3 - 240*log(3)^2 - 1200*log(3) - 2000))*log(3) + 1/4*((x^3*(log(3) + 5) + 96*x)/(log(3)^3 + 15*log(3)^2
+ 75*log(3) + 125) - 384*x/((log(3)^4 + 20*log(3)^3 + 150*log(3)^2 + 500*log(3) + 625)*x^2 - 16*log(3)^3 - 24
0*log(3)^2 - 1200*log(3) - 2000) + 240*log((x*(log(3) + 5) - 4*sqrt(log(3) + 5))/(x*(log(3) + 5) + 4*sqrt(log(
3) + 5)))/((log(3)^3 + 15*log(3)^2 + 75*log(3) + 125)*sqrt(log(3) + 5)))*log(3) + 96*(log(x^2*(log(3) + 5) - 1
6)/(log(3)^2 + 10*log(3) + 25) - 16/((log(3)^3 + 15*log(3)^2 + 75*log(3) + 125)*x^2 - 16*log(3)^2 - 160*log(3)
- 400))*log(3) - 75*x^2/(log(3)^2 + 10*log(3) + 25) + 5/4*(x^3*(log(3) + 5) + 96*x)/(log(3)^3 + 15*log(3)^2 +
75*log(3) + 125) - 480*x/((log(3)^4 + 20*log(3)^3 + 150*log(3)^2 + 500*log(3) + 625)*x^2 - 16*log(3)^3 - 240*
log(3)^2 - 1200*log(3) - 2000) + 160*x/((log(3)^3 + 15*log(3)^2 + 75*log(3) + 125)*x^2 - 16*log(3)^2 - 160*log
(3) - 400) - 20*x/(log(3)^2 + 10*log(3) + 25) - 2400*log(x^2*(log(3) + 5) - 16)/(log(3)^3 + 15*log(3)^2 + 75*l
og(3) + 125) + 480*log(x^2*(log(3) + 5) - 16)/(log(3)^2 + 10*log(3) + 25) + 300*log((x*(log(3) + 5) - 4*sqrt(l
og(3) + 5))/(x*(log(3) + 5) + 4*sqrt(log(3) + 5)))/((log(3)^3 + 15*log(3)^2 + 75*log(3) + 125)*sqrt(log(3) + 5
)) - 60*log((x*(log(3) + 5) - 4*sqrt(log(3) + 5))/(x*(log(3) + 5) + 4*sqrt(log(3) + 5)))/((log(3)^2 + 10*log(3
) + 25)*sqrt(log(3) + 5)) + 19200/((log(3)^4 + 20*log(3)^3 + 150*log(3)^2 + 500*log(3) + 625)*x^2 - 16*log(3)^
3 - 240*log(3)^2 - 1200*log(3) - 2000) - 7680/((log(3)^3 + 15*log(3)^2 + 75*log(3) + 125)*x^2 - 16*log(3)^2 -
160*log(3) - 400) + 768/((log(3)^2 + 10*log(3) + 25)*x^2 - 16*log(3) - 80)

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mupad [B]  time = 4.01, size = 139, normalized size = 5.15 \begin {gather*} \frac {x^3\,\left (\ln \left (27\right )+15\right )}{3\,\left (40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100\right )}-x\,\left (\frac {80}{40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100}-\frac {\left (128\,\ln \relax (3)+640\right )\,\left (\ln \left (27\right )+15\right )}{{\left (40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100\right )}^2}\right )-\frac {64\,x}{\left (-75\,\ln \relax (3)-15\,{\ln \relax (3)}^2-{\ln \relax (3)}^3-125\right )\,x^2+160\,\ln \relax (3)+16\,{\ln \relax (3)}^2+400}-\frac {12\,x^2\,{\left (\ln \relax (3)+5\right )}^2}{40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+100}-x^2\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6144*x + 24*x^5*log(3)^2 + exp(x)*(2048*x - log(3)*(256*x^3 + 128*x^4 - 80*x^5 - 40*x^6) + 1024*x^2 - 12
80*x^3 - 640*x^4 + 200*x^5 + 100*x^6 + log(3)^2*(8*x^5 + 4*x^6)) - log(3)*(768*x^3 - 240*x^5 + 3*x^6) - 3840*x
^3 + 80*x^4 + 600*x^5 - 15*x^6)/(4*x^4*log(3)^2 - log(3)*(128*x^2 - 40*x^4) - 640*x^2 + 100*x^4 + 1024),x)

[Out]

(x^3*(log(27) + 15))/(3*(40*log(3) + 4*log(3)^2 + 100)) - x*(80/(40*log(3) + 4*log(3)^2 + 100) - ((128*log(3)
+ 640)*(log(27) + 15))/(40*log(3) + 4*log(3)^2 + 100)^2) - (64*x)/(160*log(3) - x^2*(75*log(3) + 15*log(3)^2 +
log(3)^3 + 125) + 16*log(3)^2 + 400) - (12*x^2*(log(3) + 5)^2)/(40*log(3) + 4*log(3)^2 + 100) - x^2*exp(x)

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sympy [B]  time = 0.94, size = 75, normalized size = 2.78 \begin {gather*} \frac {x^{3}}{4 \log {\relax (3 )} + 20} - x^{2} e^{x} - 3 x^{2} + \frac {4 x}{\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 25} + \frac {64 x}{x^{2} \left (\log {\relax (3 )}^{3} + 15 \log {\relax (3 )}^{2} + 75 \log {\relax (3 )} + 125\right ) - 400 - 160 \log {\relax (3 )} - 16 \log {\relax (3 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**6-8*x**5)*ln(3)**2+(-40*x**6-80*x**5+128*x**4+256*x**3)*ln(3)-100*x**6-200*x**5+640*x**4+12
80*x**3-1024*x**2-2048*x)*exp(x)-24*x**5*ln(3)**2+(3*x**6-240*x**5+768*x**3)*ln(3)+15*x**6-600*x**5-80*x**4+38
40*x**3-6144*x)/(4*x**4*ln(3)**2+(40*x**4-128*x**2)*ln(3)+100*x**4-640*x**2+1024),x)

[Out]

x**3/(4*log(3) + 20) - x**2*exp(x) - 3*x**2 + 4*x/(log(3)**2 + 10*log(3) + 25) + 64*x/(x**2*(log(3)**3 + 15*lo
g(3)**2 + 75*log(3) + 125) - 400 - 160*log(3) - 16*log(3)**2)

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